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JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    Two plates of same mass 2 kg each are attached rigidly to the two ends of a spring (k = 100 N/m). One of the plates rests on a fixed horizontal surface and the other results some compression of the spring when it is in equilibrium state. The further minimum compression required, so that when the force causing compression is removed the lower plate is lifted off the surface will be (Take g = 10 m/s2)

    A)  0.1 m                   

    B)  0.2 m

    C)  0.3 m                   

    D)  0.4 m

    Correct Answer: D

    Solution :

     If student will try to answer it fastly he/she may calculate required extension y as \[\ge mg\Rightarrow y\ge 0.2m\]Hence option [B] is correct. Remember lower plate is lifted when upper plate is moving upward and spring is in extension condition. Compression at equilibrium \[y=\frac{mg}{k}=\frac{20}{100}=0.2\text{m}\]When extension will be equal to y then lower plate will be lifted and for this farther compression required is\[2y=2\times 0.2=0.4\,\,m\]


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